I need this for a counterexample: the multiplication in the fundamental group $\pi_1(\Sigma X_+)$, when it is equipped with the topology inherited from $\Omega \Sigma X_+$, fails to be continuous for the sort of space in the question, by a result from

2 Answers
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One of the easiest examples is the rational numbers with the subspace topology of the real line with the K-topology. Total path disconnectedness is not entirely necessary for multiplication of $\pi_{1}(\Sigma X_{+})$ to fail to be continuous. It just makes the path component space of $X$ equal to $X$, greatly simplifying complications.

Thanks, Jeremy! I knew it would be something relatively simple, but I have little intuition for point-set topology. If we drop total path disconnectedness, what do we have to assume?
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David RobertsMay 4 '10 at 22:22

Well...you have to know something about the path component space $\pi_{0}^{top}(X)$ (the quotient of $X$ where path components are identified). It is good enough to know that $\pi_{0}^{top}(X)$ is Hausdorff but not completely regular to be sure that $\pi_{1}^{top}(\Sigma X_+)$ is not a topological group. Since $\pi_{0}^{top}$ is essentially surjective this gives a counterexample for every Hausdorff, non-completely regular space. There are much better examples (locally simply connected metric spaces even!) which would probably be more interesting. These will appear in a new version soon.
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Jeremy BrazasMay 6 '10 at 12:49

There are several such examples in Steen and Seebach, Counterexamples in Topology. The first one I saw is number 60, the "relatively prime integer topology", consisting of the set $\mathbb{Z}^+$ of positive integers, with a basis of sets of the form {b + na} where (a,b)=1 and n is an integer.